Physics 207, Lecture 13, Oct. 15 - PowerPoint PPT Presentation

About This Presentation
Title:

Physics 207, Lecture 13, Oct. 15

Description:

Zero of gravitational potential energy is at bottom of the hill. R=5 m. m = 25 kg ... Skateboard. R=5 m. Physics 207: Lecture 12, Pg 12 ... – PowerPoint PPT presentation

Number of Views:17
Avg rating:3.0/5.0
Slides: 30
Provided by: michael159
Category:

less

Transcript and Presenter's Notes

Title: Physics 207, Lecture 13, Oct. 15


1
Physics 207, Lecture 13, Oct. 15
  • Agenda Finish Chapter 10, start Chapter 11
  • Chapter 10 Energy
  • Potential Energy (gravity, springs)
  • Kinetic energy
  • Mechanical Energy
  • Conservation of Energy
  • Start Chapter 11, Work
  • Assignment
  • HW5 due tonight
  • HW6 available today
  • Monday, finish reading chapter 11

2
Chapter 10 Energy
  • Rearranging Newtons Laws gives (Fd vs. ½ mv2
    relationship)
  • -2mg (yf yi ) m (vyf2 - vyi2 )
  • or ? ½ m vyi2 mgyi ½ m vyf2 mgyf
  • and adding ½ m vxi2 ½ m vzi2 and ½ m
    vxf2 ½ m vzf2
  • ½ m vi2 mgyi ½ m vf2 mgyf
  • where vi2 vxi2 vyi2 vzi2
  • ½ m v2 terms are referred to as kinetic
    energy

3
Energy
  • If only conservative forces are present, the
    total energy (sum of potential, U, and kinetic
    energies, K) of a system is conserved.

K ½ mv2 U mgy
Ki Ui Kf Uf
  • K and U may change, but E K Umech remains
    constant.

Emech is called mechanical energy
4
Another example of a conservative system The
simple pendulum.
  • Suppose we release a mass m from rest a distance
    h1 above its lowest possible point.
  • What is the maximum speed of the mass and where
    does this happen ?
  • To what height h2 does it rise on the other side ?

5
Example The simple pendulum.
  • What is the maximum speed of the mass and where
    does this happen ?
  • E K U constant and so K is maximum when U
    is a minimum.

y
yh1
y0
6
Example The simple pendulum.
  • What is the maximum speed of the mass and where
    does this happen ?
  • E K U constant and so K is maximum when U
    is a minimum
  • E mgh1 at top
  • E mgh1 ½ mv2 at bottom of the swing

y
yh1
h1
y0
v
7
Example The simple pendulum.
  • To what height h2 does it rise on the other
    side?
  • E K U constant and so when U is maximum
    again (when K 0) it will be at its highest
    point.
  • E mgh1 mgh2 or h1 h2

y
yh1h2
y0
8
Lecture 13, Exercise 1Conservation of Mechanical
Energy
A block is shot up a frictionless 40 slope with
initial velocity v. It reaches a height h before
sliding back down. The same block is shot with
the same velocity up a frictionless 20
slope. On this slope, the block reaches height
  1. 2h
  2. h
  3. h/2
  4. Greater than h, but we cant predict an exact
    value.
  5. Less than h, but we cant predict an exact value.

9
Lecture 13, ExampleThe Loop-the-Loop again
  • To complete the loop the loop, how high do we
    have to let the release the car?
  • Condition for completing the loop the loop
    Circular motion at the top of the loop (ac v2 /
    R)
  • Use fact that E U K constant !

Ubmgh
Recall that g is the source of the centripetal
acceleration and N just goes to zero is the
limiting case. Also recall the minimum speed at
the top is
Car has mass m
Umg2R
h ?
R
(A) 2R (B) 3R (C) 5/2 R (D) 23/2 R
10
Lecture 13, ExampleThe Loop-the-Loop again
  • Use E K U constant
  • mgh 0 mg 2R ½ mv2
  • mgh mg 2R ½ mgR 5/2 mgR
  • h 5/2 R

h ?
R
(A) 2R (B) 3R 5/2 R (D) 23/2 R
11
Lecture 13, ExampleSkateboard
  • What speed will the skateboarder reach at bottom
    of the hill if there is no friction and the
    skeateboarder starts at rest?
  • Assume we can treat the skateboarder as point
  • Zero of gravitational potential energy is at
    bottom of the hill

m 25 kg
R5 m
R5 m
12
Lecture 13, ExampleSkateboard
  • What speed will the skateboarder reach at bottom
    of the hill if there is no friction and the
    skeateboarder starts at rest?
  • Assume we can treat the skateboarder as point
  • Zero of gravitational potential energy is at
    bottom of the hill

m 25 kg
  • Use E K U constant
  • Ebefore Eafter
  • 0 mgR ½ mv2 0
  • 2gR v2 ? v (2gR)½
  • v (2 x 10 x 5)½ 10 m/s

R5 m
R5 m
13
Potential Energy, Energy Transfer and Path
  • A ball of mass m, initially at rest, is released
    and follows three difference paths. All surfaces
    are frictionless
  • The ball is dropped
  • The ball slides down a straight incline
  • The ball slides down a curved incline
  • After traveling a vertical distance h, how do the
    three speeds compare?

1
3
2
h
(A) 1 gt 2 gt 3 (B) 3 gt 2 gt 1 (C) 3 2 1
(D) Cant tell
14
Lecture 13, Exercise 2 Potential Energy, Energy
Transfer and Path
  • A ball of mass m, initially at rest, is released
    and follows three difference paths. All surfaces
    are frictionless
  • The ball is dropped
  • The ball slides down a straight incline
  • The ball slides down a curved incline
  • After traveling a vertical distance h, how do the
    speeds compare?
  1. 1 gt 2 gt 3
  2. 3 gt 2 gt 1
  3. 3 2 1
  4. Cant tell

15
Potential Energy, Energy Transfer and Path
  • A ball of mass m, initially at rest, is released
    and follows three difference paths. All surfaces
    are frictionless
  • The ball is dropped
  • The ball slides down a straight incline
  • The ball slides down a curved incline
  • After traveling a vertical distance h, how do the
    three speeds compare?

(A) 1 gt 2 gt 3 (B) 3 gt 2 gt 1 (C) 3 2 1
(D) Cant tell
16
Elastic vs. Inelastic Collisions
  • A collision is said to be elastic when energy as
    well as momentum is conserved before and after
    the collision.
    Kbefore Kafter
  • Carts colliding with a perfect spring, billiard
    balls, etc.
  • A collision is said to be inelastic when energy
    is not conserved before and after the collision,
    but momentum is conserved.
  • Kbefore ? Kafter
  • Car crashes, collisions where objects stick
    together, etc.

17
Inelastic collision in 1-D Example 1
  • A block of mass M is initially at rest on a
    frictionless horizontal surface. A bullet of
    mass m is fired at the block with a muzzle
    velocity (speed) v. The bullet lodges in the
    block, and the block ends up with a speed V.
  • What is the initial energy of the system ?
  • What is the final energy of the system ?
  • Is energy conserved?

x
v
V
before
after
18
Inelastic collision in 1-D Example 1
  • What is the momentum of the bullet with speed v
    ?
  • What is the initial energy of the system ?
  • What is the final energy of the system ?
  • Is momentum conserved (yes)?
  • Is energy conserved? Examine Ebefore-Eafter

v
No!
V
x
before
after
19
Example Fully Elastic Collision
  • Suppose I have 2 identical bumper cars.
  • One is motionless and the other is approaching it
    with velocity v1. If they collide elastically,
    what is the final velocity of each car ?
  • Identical means m1 m2 m
  • Initially vGreen v1 and vRed 0
  • COM ? mv1 0 mv1f mv2f ? v1 v1f
    v2f
  • COE ? ½ mv12 ½ mv1f2 ½ mv2f2 ? v12 v1f2
    v2f2
  • v12 (v1f v2f)2 v1f2 2v1fv2f v2f2 ? 2
    v1f v2f 0
  • Soln 1 v1f 0 and v2f v1 Soln 2 v2f
    0 and v1f v1

20
Lecture 13, Exercise for homeElastic Collisions
  • I have a line of 3 bumper cars all touching. A
    fourth car smashes into the others from behind.
    Is it possible to satisfy both conservation of
    energy and momentum if two cars are moving after
    the collision?
  • All masses are identical, elastic collision.
  • (A) Yes (B) No (C) Only in one special
    case

v
Before
v1
v2
After?
21
Lecture 13, Exercise for homeElastic Collisions
  • COM ? mv mv1 mv2 so v v1 v2
  • COE ? ½ mv2 ½ mv12 ½ mv22
  • v2 (v1 v2)2 v12 v22 ? v1 v2 0
  • (A) Yes (B) No (C) Only in one special
    case

Before
After?
22
Variable force devices Hookes Law Springs
  • Springs are everywhere, (probe microscopes, DNA,
    an effective interaction between atoms)
  • In this spring, the magnitude of the force
    increases as the spring is further compressed (a
    displacement).
  • Hookes Law,
  • Fs - k Dx
  • Dx is the amount the spring is stretched or
    compressed from it resting position.

Rest or equilibrium position
F
Dx
23
Hookes Law Spring
  • For a spring we know that Fx -kx.

24
Lecture 13, ExampleHookes Law
  • Remembering Hookes Law, Fx -k Dx
  • What are the units for the constant k ?
  • (A) (B) (C) (D)

F is in kg m/s2 and dividing by m gives kg/s2 or
N/m
25
Lecture 13, Exercise 2Hookes Law
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
26
Lecture 10, Exercise 2Hookes Law
8 m
9 m
What is the spring constant k ? SF 0 Fs
mg k Dx - mg Use k mg/Dx 5 N / 0.01 m
Fspring
50 kg
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
27
F-x relation for a foot arch
Force (N)
Displacement (mm)
28
F-x relation for a single DNA molecule
29
Measurement technique optical tweezers
30
Lecture 13, Oct. 15
  • Chapter 10 Energy
  • Potential Energy (gravity, springs)
  • Kinetic energy
  • Mechanical Energy
  • Conservation of Energy
  • Chapter 11, Work
  • Assignment
  • HW5 due tonight
  • HW6 available today
  • Monday, finish Chapter 11
Write a Comment
User Comments (0)
About PowerShow.com